標題:
maths question 7and8
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7)(a) Determine the remainder when 2^2009+1 is divided by 17 (b) Prove that 30^99+61^100 is divided by 31 (c) It is known that numbers p and 8p^2+1 are primes.Find p.8) Digits 2,3,4,,5 and 7 are each used once to compose a 5-digits number abcde such that 4 divides a 3-digits number abc , 5 divides a... 顯示更多 7)(a) Determine the remainder when 2^2009+1 is divided by 17 (b) Prove that 30^99+61^100 is divided by 31 (c) It is known that numbers p and 8p^2+1 are primes.Find p. 8) Digits 2,3,4,,5 and 7 are each used once to compose a 5-digits number abcde such that 4 divides a 3-digits number abc , 5 divides a 3-digits numder bcd and 3 divides a 3-digits number cde. Find the 5-digits numder abcde. please show the steps
最佳解答:
7) Consider binomial theorem (a+b)^n = (nC0)a^n + (nC1)a^(n-1)b+ ... (nCn)b^n, all terms are divisible by a except for the last term (nCn)b^n. (a)2^2009+1 = 2*2^2008+1 = 2*(2^4)^502 + 1 = 2*16^502+1 = 2*(17-1)^502 + 1 Note that in the binomial expansion of (17 - 1)^502, all terms are divisible by 17 except for the last term (-1)^502 So 2*(17-1)^502 + 1 can be rewritten as 2[17K + (-1)^502] + 1 where K is an interger. As such the remainder when 2^2009+1 is divided by 17 is 2*(-1)^502 + 1 is 3 (b) Rewriting 30^99 + 61^100 as (31 - 1)^99 + (62-1)^100 Applying same reasoning as in (a), all terms of the 2 binomial expansions are divisible by 31 and 62 respectively excpet the last terms (-1)^99 and (-1)^100. So 30^99+61^100 = 31K + (-1)^99 + 62H + (-1)^100 where K and H are integers. = 31(K+2H) -1 + 1 = 31(K+2H) is divisible by 31 (c) It is known that numbers p and 8p^2+1 are primes.Find p. All positive integers can be expressed as 3k-2, 3k-1 and 3k where k = 1,2,3,4... when p=3k-2, 8p^2+1 = 8*(3k-2)^2 + 1 = 8*(9k^2 - 12k + 4) + 1 = 72k^2 - 96k + 33 is divisible by 3 (not prime) when p=3k-1, 8p^2+1 = 8*(3k-1)^2 + 1 = 8*(9k^2 - 6k +1) + 1 = 72k^2 - 48k + 9 is divisible by 3 (not prime) when p =3k, p is not a prime except when k=1 or p=3 8*3^2+1 = 73 which is the only solution to the question. 8) A = abc divisble by 4 => (i) c= 2 or 4 (even) and (ii) bc divisible by 4 B = bcd divisible by 5 => d = 5 C = cde or c5e divisible by 3 => c + 5 + e is divisible by 3 if c=2, cannot find an e from [3,4,7] to make 2 + 5 + e divisible by 3 if c=4, can only find e=3 from [3,4,7] to make 4 + 5 + e divisible by 3. So c=4, e=3 Since b4 is divisible by 4, only b=2 from [2,7] is possible. So b=2, a=7 The 5 digit number is 72453
其他解答:
7(a) 2^4=-1 (mod 17) 2009=4*502+1 2^2009=-1 (mod 17) So 2^2009+1=-1+1=0 (mod 17) The remainder is 0 (b) By Euler theorem 30^30=61^30=1 (mod 31) 30^99+61^100 =30^9+61^10 (mod 31) =(-1)^9+(-1)^10 (mod 31) =-1+1 (mod 31) =0 (mod 31) So 30^99+61^100 is divided by 31 (c) p and 8p^2+1 are primes Notice that when p=3,8p^2+1=73 which is a prime if p not equal to 3, then p=+1 or -1 (mod 3) =>8p^2+1=0 (mod 3) shows that 8p^2+1 is composite. We conclude that p=3 8 abcde where 4|abc,5|bcd,3|cde [a,b,c,d,e come from 2,3,4,5,7] So d is 5,3|c+e+5 c+e=7,10 On the other hand 4|bc =>c=4,e=3,b=2,a=7 The number is 72453
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